Examples: $2$ & $3$, $3$ & $5$, $5$ & $7$, $59$ & $61$, $99999989$ & $100000007$.
I was inspired by the fact that, in English, all consecutive numbers share a common letter, such as seven & eight that share 'e'. See Alex Bellos's Math Puzzle section https://www.theguardian.com/science/2023/feb/20/did-you-solve-it-thats-mathematics. Instead I consider consecutive primes and their base representation.
Except for $2$ & $3$, $3$ & $5$ and $5$ & $7$ all such base-$10$ consecutive primes are of the form: $a \times 10^n - b$ and $a \times 10^n + c$ with $1 \le a\le 8$. It is easy to show that $a$ cannot be $9$. I found $701$ cases below $10^{500}$. A large example is $1 \times 10^{101}-203\ \ $ & $\ \ 1 \times 10^{101}+3$ and a larger example (with probable primes) is $8 \times 10^{5002}-6243\ \ $ & $\ \ 8 \times 10^{5002}+14481$.
I am pessimistic that the conjecture can be easily proved. Landau's conjecture that there are infinitely many primes $p$ of the form $p=n^2+1$ remains open.
For other bases: in base $2$ there are no such consecutive primes. In base $3$ it is easy to show that there are none, except $2$ & $3$. Base $4$ is the first interesting base. All cases must be of the form $4^n - b$ and $4^n + c$ (except for $2$ & $3$). I found only $5$ cases with $n$ = $1$, $4$, $28$, $83$ and $1816$.
EDIT: Replaced "disjoint" with the more standard "have no common digits". See the result for squares in oeis.org/A156981
Heuristically, we expect the typical gap between primes of $n$ digits to be on the order of $n$. When this is much larger than $10^{10}$, it becomes unlikely that the greatest prime $< a \times 10^{n}$ and the least prime $> a \times 10^{n}$ are disjoint. So I would suspect that there are only finitely many disjoint consecutive primes. Of course this is not a proof.